If $ f:\mathbb R^n\to \mathbb R $ be bounded and continous then differential equation $$x'=f(x)$$ has a solution with domain $\mathbb R$.
outlook of proof : if the maximal domain of solution is $(a,b)$ and $b<\infty$ then we see that $$|x(t_m)-x(t_n)|=|\int_{t_n}^{t_m} f(x(s)) ds |\le M|t_n-t_m|$$ for every sequence $t_n \to b$ so we can extend $x(t)$ continously to $b$ and then easly we can show that $b$ is contained in the maximal domain so $b=\infty$ and similarly $a=-\infty$
I think if $ f:\mathbb R^n\to \mathbb R $ be uniformly continous then we can have solution for differential equation above with maximal domain $\mathbb R$.
My attempt to proof this is :
we can show if $ f:\mathbb R^n\to \mathbb R $ be uniformly continous there exists $A$ and $B$ such that $|f(x)| \le A|x|+B $ and so similarly if maximal domain is $(a,b)$ with $b<\infty$ we have $$|x(t_m)-x(t_n)|=|\int_{t_n}^{t_m} f(x(s)) ds |\le \int_{t_n}^{t_m} (A|x(s)|+B)ds $$ now with the help of Generalized gronwall inequality we can see that $$ |x(t)|\le (Bt)(e^{At})) $$ and because the solution $|x(t)|$ is bounded near $b$ and so we an extend continuously solution to $b$ and do similarly like above proof.
Can someone see that this is true or not?
Yes, it is true. In fact, as your proof shows, the inequality $|f(x)|\le A\,|x|+B$ is enough.